materials notes

7/28/2019 Materials Notes

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MATS1101 - Materials - Bonding Between Atoms

Introduction to Materials Engineering

Why Study Materials?

To understand, predict and control material behaviour.

Enables us to use the correct materials for a given application and understand the

limitations on their use.

To determine material behaviour we must understand the material properties, which are aresult of the internal structure of the material. We need therefore to understand thestructure/property relationship.

Internal Structure-determines->Material Properties-dictates->Material Behaviour

Atomic Bonding

All materials are made up of atoms, and these atoms are held together by forces (calledinteratomic bonding). These forces act like springs, linking each atom to its neighbour.

These atoms are arranged in different ways in different materials. The two importantthings to understand for atom packing are:

Number of bonds per unit area.

The angle of the bonds.

There are two different ways that atoms can be bound together.

Primary bonds. These are strong bonds and may be ionic, covalent or metallic.

Secondary bonds. These are weak bonds and may be Van der Waals or hydrogen.

Note: Although these different bonds are distinguished, in many materials there exists amixture of different bonds.

Primary Bonds

All elements except inert gases have an unfilled outer shell of electrons (valenceelectrons).

Primary bonding occurs when electrons are lost or gained so that the outer shell is filled.

Ionic BondingThe ionic bond forms between two elements, one with a small number of electrons in the

valence shell (metal) and one with an almost full outer shell (non-metals).

Na and Cl form an ionic bond; with Na giving up an electron from its valance shell and

donating it to the Cl atom to complete its valence shell.

The non-metal atoms attract valence electrons from metal atoms to become

negatively charged, e.g. chlorine accepts an electron to become Cl -.

The metal atom releases electrons to become positively charged, e.g. sodium loses

an electron to become Na

+

.

These charged atoms (ions) are then held together by attraction of opposite chargescreating an ionic bond between them.

Ionic bonds are not directional, because the bonding is created by opposite charges onthe ions. This means the ions have freedom in the way that they pack. However, ions ofopposite sign must surround each other to retain the attraction between ions.

Ionic bonds are:

Strong and stiff

Give a material high strength

High elastic modulus

High melting point

Poor electrical conductivity

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Ionic bonding is the dominant bonding in some ceramics such as magnesia (MgO),alumina (Al2O3) and cement.

Covalent BondingIn covalent bonding electrons are shared between neighbouring atoms. E.g. in a

diamond, each carbon atom shares its valence electrons with four other carbon atoms.A 3D structure is formed with covalent bonding with the bonds being very directional

because the electrons are shared between atoms. Directionality of bonds dictates the atompacking. In a diamond, the carbon atoms build up into a 3D array with the bonds pointingtowards the corners of tetrahedron.

Covalent bonds are:

Very stiff

Give very high elastic modulus

High (inherent) strength

High melting point

Low electrical conductivity

Covalent bonding is the dominant bonding found in silicate ceramics and glasses. Theyalso occur in the backbone of polymer chains and in the cross-links in thermosetting

polymers.

Metallic BondingIn metallic bonding the electrons are surrendered to common pool and become shared

by all the atoms in the solid metal.When the atoms give up their electrons, they become positively charged and these ions

are then held firmly in place by the attractive forces between the ions and the negativelycharged electron cloud. Bonds are not directional, so atoms tend to pack to give simple,dense structures.

The metallic bond is:

Strong and stiff

Giving high elastic modulus

High strength

Good electrical conductivity (because electrons have easy movement) Metallic lustre

Good ductility

Secondary Bonds

Secondary bonds are created from weak attraction present due to asymmetrical chargedistribution in adjacent atoms.

Van der Waals BondingThis is a weak attraction caused by a temporary dipole between unchanged atoms.

Although overall there is no charge on an atom, as the electrons whiz around in their orbit the

atom is instantaneously negative on the side where the electron is situated. Theinstantaneous dipole created by one atom creates a dipole on a nearby atom and anattractive force is created. This is the bonding present in N2 molecules.

Hydrogen BondingThis a weak bonding due to a dipole produced by hydrogen when it is covalently bonded

to other atoms. The hydrogen becomes slightly more positive than the atom it is bonded to,and therefore a dipole is produced.

An example of hydrogen bonding is ice, where H2O molecules has slight dipole becausethe oxygen atom is slightly negative compared to the hydrogen.

The hydrogen bond is the strongest type of secondary bond.

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Bond Energy

Atoms experience repulsive forces as well as attractive forces, because as the atoms getcloser together, the attraction only continues until the charge distribution starts to overlap. Theequilibrium spacing of atoms occurs where the bond energy is a minimum, i.e. where the netcurve of attraction and repulsion has the minimum value.

The force between atoms, F, is given by:

dr

dU

where U = bond energy and r = atomic separation

Note that the force is zero at the equilibrium spacing; if atoms are pulled apart a smalldistance there is a resisting force.

Bond stiffness, S, is given by:

2

2

dr

Ud

dr

dFS ==

When the stretching is small, S is constant and therefore the bond is linear-elastic. Thismeans that as the interatomic distances increases, the force required increases linearly.

Bond force determines the:

Elastic modulus (or Youngs modulus) of a material (how stiff a material is)

Melting or softening temperature (stronger bonds will enable a material to withstand

higher temperature before bonds break)

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MATS1101 - Materials - Packing of Atoms in Solids

Atom Packing

For most engineering materials, atoms pack in regularly repeating 3D arrays. These 3Darrays create crystalline grains which are assembled together to make up the solid.

Atom->3D Arrays->Grains->Solid

Different materials have different types of packing, depending on the: Type of bonding

The sizes of atoms involved and

The different elements present.

Metal Crystal Structures

Bonds are non-directional in metals, atoms pack together efficiently into close packed ornearly close packed structures.

Body Centred Cubic (BCC)Each atom is surrounded by 8 other atoms, each of which is at the corner of a cube. The

number of nearest neighbours, which is termed the co-ordination number, is 8. E.g. iron.

Face Centred Cubic (FCC)There is one atom at each corner of the cube and an atom at the centre of each face of

the cube. The co-ordination number is 12. E.g. aluminium, copper and nickel.

Close Packed Hexagonal (CPH)Atoms are arranged at corners of a hexagonal prism with an atom at the centre of the top

and bottom face and three atoms within the prism. The co-ordination number is 12. E.g.titanium, zinc and magnesium.

Ceramic Crystal Structures

There are more than one species of atom in ceramics, e.g. aluminium and oxygen inalumina, which makes most ceramic crystal structures quite complicated.

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Polymer Structure

Polymers consist of huge molecules, forming spaghetti-like chains, held together bycovalent bonding. The chains are generally arranged randomly, i.e. the chains are notarranged in a regular repeating pattern. This means that polymers are non-crystalline. Theterm amorphous is used to describe non-crystalline materials.

Glass Structure

Glasses are also amorphous since they also have no regular arrangement of the atoms.

Density

Density of materials depends on the size, mass and efficiency of packing of the atoms.Metals are made up of heavy atoms that are densely packed; therefore they have a highdensity. Polymers and ceramics are made up of light atoms that are generally less closelypacked; therefore they tend to have lower density.

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MATS1101 - Materials - Youngs Modulus

The Elastic Moduli

The elastic modulus measures the resistance of the material to elastic or springydeformation. Low modulus materials are floppy they suffer large deflections if they areloaded. Low modulus materials are desirable for springs, cushions or vaulting poles.

For most engineering applications, a high modulus material is desired to withstand high

loads with minimum deflection.Hookes Law is a description of the observation from experiments that when strain is

small, the strain is proportional to the stress. From this law, the elastic modulus is defined asthe ratio of stress, , to strain, , at small strains, i.e.

= Ewhere E = Youngs Modulus.

This law holds for both tension and compression loading.In the same way, shear strain, , is related to shear stress, , by shear modulus, G, i.e.

= G

Pressure causes a shrinkage of volume, and so the negative of dilatation is proportionalto the pressure. Bulk modulus is therefore defined as:

p = -Kwhere K = bulk modulus, p = pressure and = dilatation

Poissons ratio, u, is the ratio of lateral shrinkage strain to longitudinal tensile strain, i.e.lateral strain = - tensile strain

E, G, K and are the commonly used elastic constants.

Physical Basis of Elastic Behaviour

Bonding between AtomsAtoms in crystals are held together by bonds that behave like little springs. Atoms

experience both attractive and repulsive forces. Atoms displaced from the equilibrium

position, r0, create restoring forces, pulling the atom back to the equilibrium position. This isthe origin of elasticity.

Stiffness, S, of a bond is given by:dr

dFS =

When stretching is small, S is constant at S0, where0

0

rrdr

dFS

=

=

This means that the bond behaves in a linear elastic way and for small displacements:F = S0 (r-r0)

Consider unit area of material containing N bonds, stretched as shown:

The total force exerted across the unit area is the stress, .

N is the number of bonds per unit area. Since the atom spacing = r0, the average area peratom is r0

2. Therefore:

2

0

1

rN =

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MATS1101 - Materials - Youngs Modulus

Stress = force / unit area= (no. of bonds / unit area) x stiffness x r

2

0

0

000

)()(NS

r

rrSrr

==

As strain0

0

r

rr

E

= , Youngs Modulus

0

0

r

S

E ==

S0 can be calculated from theoretical bond energies, allowing calculations of E.

These calculations of E agree well with the moduli of metals and ceramics. Howeverpolymers generally have a lower modulus. This is because for polymers, the elastic behaviouris dominated by the secondary Van der Waal's bonds, between the spaghetti-like polymerchains rather than the covalent bonds in the backbone, and polymers therefore have lowstiffness. In some polymers, such as polyethylene, the secondary bonds melt below roomtemperature and the stiffness (at room temperature) is even lower. The same is true forrubbers.

In some polymers, some covalent bonds form between the polymer chains as well as thesecondary bonds. Like the secondary bonds, these also contribute to stiffness, so that thestiffness increases as the amount of covalent cross-linking increases.

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MATS1101 - Materials - Yield and Tensile Strength

Yield and Tensile Strength

There are many engineering situations where yield strength becomes an important designconsideration. Engineers generally aim to design against permanent deformation in structures(yield limited design).

All solids have an elastic limit, beyond which the following occurs, depending on the

material:1. Brittle materials undergo fracture, e.g. glass (fractures suddenly) and concrete

(cracks gradually).2. Ductile materials undergo plastic deformation (i.e. change shape permanently), e.g.

most metal alloys and elastomers will deform permanently when stretched too far.This is also known as plastic collapse.

The strength at the elastic limit is the Yield Strength.

Characterisation of StrengthThe tensile test is used to measure a materials strength. The test piece has a measured

gauge length marked off on the reduced section. The test piece is usually circular in cross-section, but rectangular specimens are also used.

A force is applied longitudinally to the test piece in the tensile test and measurements are

made based on the elongation of the material under the applied force. Parameters are givenbelow:

L = Gauge lengthD = Diameter of the test pieceF = Force applied to the test pieceA = Cross sectional area of the test piece = D2

n = Nominal stress = F / A0 (A0 is the original cross sectional area)n = Nominal strain = (L-L0) / L0 = L / L0 (L0 is the original gauge length, L is the

distance between gauge marks after the load has been applied.

The force is plotted against the extension of the test piece to give a load-extension curve.

As the test progresses, increasing force is applied to the test piece. Initially the test pieceexhibits linear elastic behaviour, i.e. the extension increases linearly with the load, until theelastic limit. After this point, the test piece yields and continues to elongate uniformly until theload reaches a maximum point. This load is used to define the Ultimate Tensile Strength(UTS), which is the nominal stress at the maximum load. After the maximum load, the testpiece necks (called plastic instability) and there is non-uniform elongation (the pieceelongates in a localised region) until the test piece breaks.

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The nominal stress and strain are determined from the load-elongation curve and areplotted (n vs. n) to give an engineering stress-strain curve. This curve gives a number ofparameters used for material characterisation. This curve is used because the results of theengineering stress-strain curve apply to all sizes and shapes of test pieces.

y = Yield Strength = Stress at the elastic limit (at the onset of plastic flow). It is the stressthat divides the elastic and plastic behaviour of the material. To design a part that will notdeform plastically in service, we must use a material with a high yield strength or make thecomponent large enough that the force produces a stress below the yield strength.

0.1% = 0.1% Proof Stress = Stress at a permanent strain of 0.1%.Often 0.2% Proof Stress is also used.Proof Stress is used for materials that yield gradually and therefore do not exhibit a

distinct elastic limit or yield point. A line is drawn parallel to the elastic region of the stress-strain curve, offset by 0.001 (or 0.002). The proof stress is the stress at which this lineintersects the stress-strain curve.

TS = Tensile Strength or Ultimate Tensile Strength = Stress at maximum load or onset ofnecking.

E = Young's Modulus or Modulus of Elasticity = slope of the stress-strain curve within theelastic region. The modulus gives an indication of a material's stiffness or resistance to elasticdeformation.

f = Plastic Strain After Fracture or Tensile Ductility = Permanent elongation resulting fromthe tensile test. The broken pieces are put together and final gauge length l f is measured.Strain f = (lf-l0)/l0, often called percent elongation (if multiplied by 100).

The ductility of the material is important in design, where ductility is needed so that thepart deforms before fracturing, and in manufacturing, where ductility is required to formcomplicated shapes without breaking the material.

The energy expended in deforming a material per unit volume is given by the area underthe stress-strain curve.

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Linear ElasticityMost solids show linear elastic behaviour at small strains (


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MATS1101 - Materials - Dislocations

Dislocations

Ideal Strength of MaterialsThe slope of the interatomic force-distance curve at equilibrium spacing is proportional to

Youngs Modulus, E.Interatomic forces become negligible for r > 2r0.

The maximum in the force-distance curve occurs at ~1.25r0, (where F = Fmax). If appliedstress is greater than Fmax per bond, bonds between atoms are broken and fracture occurs.Ideal strength, , corresponds to bond rupture at Fmax.

Calculation of Ideal StrengthFrom the force-distance curve, where r = 2r0, = 0.25 and 2, ( is ideal strength).

A better estimate using interatomic potential gives

Glasses and some ceramics have a yield strength of

For other ceramics and polymers,

For metals,

Actual vs. Ideal StrengthGlasses and some ceramics show strengths close to their ideal strength. Polymers are

also close to their ideal strength - although their strength is low, their modulus is also low.Why is y < for metals and some ceramics? Crystals are not perfect and contain

defects. One of these defects is a dislocation. In these materials that do not exhibit idealstrengths, plastic (permanent) deformation occurs by shear stress at stresses much less than

the ideal strength, via a dislocation mechanism.

Yield occurs by shear stress because tensile and compressive stresses create shearstresses. The shear stress is at a maximum at 45 to the axis of tension or compression.

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MATS1101 - Materials - Dislocations

Dislocation Mechanism of Plastic Deformation

Shear deformation occurs by creation and motion of dislocations on crystal planes.Dislocations are carriers of deformation as electrons are carriers of charge. They exist inalmost all crystalline solids, typically introduced during solidification or deformation.Dislocations are line defects in the crystal structure, creating lattice distortion around the

dislocation. Atoms are rearranged as the dislocation moves through the crystal. A dislocationis shown below:

Dislocations move easily (glide) at low stress because:1. Atom displacements are small2. Atom displacements are localised near the dislocation line (i.e. only a small fraction of

atoms are displaced at any time).

Dislocations form by mis-stacking of atoms during crystallisation. Typical engineeringalloys contain ~100 000 km of dislocation lines per cm3.

Dislocation glide is easy in metals due to the non-specific nature of metallic bonding.

However it is difficult in ceramics due to the specific nature of covalent or ionic bonding.

With covalent bonding the strength and directionality of the bonds prevent

dislocations moving.

With ionic bonding, movement of the dislocation disrupts the charge balance around

surrounding atoms.

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MATS1101 - Materials - Strengthening Methods

Strengthening Methods

Forces on DislocationForce, f, resisting motion in slip plane:

f = .b = shear stress

b = burgers vector

Force, T, along dislocation line (line tension):T = Gb2

G = shear modulus

A crystal yields when force on dislocation, b, is greater than the force opposing motion, f.Thus, at yielding, dislocation yield strength, y, is related to lattice resistance to dislocationmotion, f, by:

Intrinsic Lattice Resistance to Dislocation Glide

The basic component of lattice resistance f is fi, the intrinsic lattice resistance - this iscaused by the bonds between atoms that have to be broken and reformed as the dislocationmoves.

In metals, fi is low due to the non-specific nature of metallic bonding, e.g. pure singlecrystals. In ceramics, fi is high due to the specific nature of covalent or ionic bonding, e.g.diamond. As metals have a low fi, it is useful to be able to increase f in metals bystrengthening.

The overall lattice resistance, f, of metal crystals may be increased by:

solution strengthening: fss

precipitates (obstacles): fo

work hardening: fwh

because the effect of these adds to fi.

The upper limit on strengthening is always less than or equal to the ideal strength. Inpractice, the ideal strength is rarely approached.

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Solid Solution StrengtheningImpurity atoms dissolve in solid metal to give an alloy. An example is brass, where zinc

atoms replace copper atoms randomly, up to approximately 30% zinc. Impurity atoms distortthe slip planes, because of atomic size mismatch and therefore dislocation motion isimpeded.

y is increased by fss/b

The degree of solid solution strengthening depends on two factors:1. The larger the size difference between the two atoms, the greater the strengthening.2. The amount of impurity atoms added. fss is increased as the impurity concentration

increases. Since impurity atom spacing on the slip plane is proportional to C -;y C

Precipitation and Dispersion StrengtheningImpurity atoms dissolved at high temperature may precipitate on cooling. This creates

fine, closely spaced precipitates that impede dislocation motion. An example is Al-Cu (2000Aluminium Series) alloys, HSLA steels and precipitation hardened (p-h) stainless steels.Small precipitates or particles or particles obstruct the motion of dislocations, as shownbelow:

For the critical situation (c), the force bL on the segment of dislocation between twoadjacent particles = force 2T due to line tension, i.e.

Obstacles exert resistance on dislocationsMaximum effect is found with strong, closely-spaced particles (high T, low L).

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Work HardeningMetal crystals (grains) have several slip planes. When a stress is applied to a material,

dislocations are created in the material. The dislocations moving along intersecting slip planesinteract - they obstruct one another and accumulate. The more dislocations there are, themore likely they are to interfere with each other and the stronger the material becomes. Theyare only removed by annealing (heating).

The stress-strain curve rises after the yield point and the increase can be described by:y

m

where m is the strain hardening index

The lattice resistance increases by fwh (usually substantial). Examples of work hardeningalloys are Al-Mn (3000 Aluminium alloys), brass, bronze and stainless steel.

Strengthening contributions are additive. Other strengthening mechanisms are alsopossible, for example, transformation strengthening in hardened steels and orderstrengthening in superalloys.

Strong materials either have high fi, e.g. diamond, or they use other mechanisms, e.g.high-strength alloys. y is yield strength of isolated grains in shear, but we need the yieldstrength of polycrystalline materials in tension.

Grains have different orientations within a material. Yielding occurs progressively fromgrain to grain and is constrained by adjacent grains. Slip begins in grains where there are slipplanes nearly parallel to , e.g. grain 1. Slip will then spread to grains like grain 2 and lastly tograins like grain 3.

The yielding takes place progressively in polycrystalline materials, which is why they don'thave a sharp yield point on the stress-strain curve. The yield strength is therefore higher inpolycrystalline materials than if the material was just a single grain. For polycrystallinematerials y increases by Taylor factor of 1.5. The stress conversion factor is = 2. Thereforepolycrystalline materials yield when y = 2 1.5 y. But grain size also affects y.

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Grain Size StrengtheningGrain boundaries (GBs) are surface defects in the crystal structure.

Grain boundaries impede slip dislocations and dislocations pile up at grain boundaries,creating a back-stress on dislocations. Therefore the dislocation yield stress, y, increases.Grain boundary spacing in slip planes decreases as the grain size decreases, therefore asgrain size decreases, y increases.

In low-carbon steels, the Hall-Petch equation holds:y = o + const x d-

0 is related to yd = average grain diameter

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MATS1101 - Materials - Fast Fracture

Fast Fracture

Role of CracksCatastrophic failure by fast fracture may occur in structures even though they are properly

designed to avoid yielding and excessive elastic deformation, e.g. welded ships, pressurevessels, welded gas pipelines, welded bridges. In situations where unexpected failure occurs,

the common feature present is usually unstable cracks.

Energy for Crack GrowthWhen cracks are present, there is a critical stress that must be exceeded for the crack to

grow. To calculate this critical stress, we examine the work required for crack growth. Thecriterion for crack advancement is that the work done by the load must be greater than thedifference between the change of elastic energy and the energy absorbed at the crack tip.

For a crack of length a, which advances a distance a, in a plate of thickness t, under aload F:

Work done by load = WChange of elastic energy = Uel

Energy absorbed at crack tip = Gc.t.aW Uel + Gc.t.a

where Gc is energy absorbed per unit area of crack (not only surface energy) and t. a isnew crack area

ToughnessGc is a material property that indicates the strain energy release rate, or "toughness" of a

material. It is measured in energy per unit area, with units of J.m-2.

A high Gc value indicates that crack propagation is difficult in that material, e.g.

copper is tough and has a Gc of approximately 106 J.m-2.

A low Gc value indicates that crack propagation is easy in that material, e.g. glass is

not tough and has a Gc of approximately 10 J.m-2.

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Stress for Crack GrowthConsider a plate with fixed ends, under tension, e.g. welds in steel structures.

The ends cannot move and therefore the forces acting do no work i.e. W = 0 andtherefore:

-Uel = GctaAs the crack advances, the material relaxes - elastic strain energy is lost, i.e. Uel is

negative, and Gc is positive.

Inside the plate, for a unit cube element under stress, , and strain, , strain energy is:

Strain energy released per unit volume in the shaded region due to the crack is:

Overall energy change in the shaded region due to crack is:

If crack grows by a:

The critical condition for fixed displacement:-Uel = Gcta

then gives:

However, the assumption about relaxation under-estimates Uel by about 2, so

Rearranging gives:

at onset of fast fracture

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Consider a plate with a fixed load. As the crack grows, the plate relaxes and becomesless stiff. Applied forces move and do work, i.e. W is positive. However Uel is also positivebecause some of W is used to increase Uel and the final result is the same.

The equation derived for both these situations is:

at onset of fast fracture

The right hand side is constant for a given material because these are material propertiesonly.

The left hand side gives either:

the critical value of at which the crack of length a will propagate or

the critical crack length a, which will become unstable under a given stress, , and

propagate with fast fracture.

Fracture ToughnessDefining a stress intensity factor, K (MN.m-3/2), where

Fast fracture occurs when K = Kc, where Kc is critical stress intensity factor, or fracture

toughness, given by:

Combining this with the definition of stress intensity factor, the fast fracture conditionbecomes:

It can be seen that:

Ceramics have the lowest Gc and the lowest Kc.

Metals have the highest Gc and the highest Kc.

Polymers have an intermediate Gc and a low Kc (because of a low E).

Composites have a higher Gc than polymers and an intermediate Kc, also greater than

polymers.

Note: some materials have a transition in behaviour with temperature, eg. bcc metals likesteel becomes quite brittle if cooled sufficiently.

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MATS1101 - Materials - Micromechanisms of Fracture

Micromechanisms of Fracture

Crack Propagation ResistanceThe toughness of materials varies:

Metals: high Gc and Kc, which means that cracks propagate slowly

Ceramics: low Gc and Kc, which means that cracks propagate fast

Polymers: intermediate Gc and low Kc

Composites: Gc and Kc greater than for polymers

These materials are different because they have different stress concentration effects atthe crack tip and different micromechanisms of fracture.

Stress ConcentrationIn a material near the crack, local > , where is the average stress.As distance from the crack tip, r, decreases, local increases, because:

Approaching the crack tip, a distance, ry, is reached where local approaches y. Plasticflow then occurs, forming a plastic zone at the crack tip.

To find ry, we set local = y and assume ry is much smaller than the crack length a:

since:

The crack will grow when K = Kc and the size of the plastic zone is given by the definitionof ry above.

From the above equation, as y increases, the plastic zone size, ry, decreases. Thereforestrong materials will have a small plastic zone, or none, e.g. ultra-high strength alloys orceramics. Low strength materials will have large plastic zones, e.g. pure metals and ductilealloys.

Ductile RuptureDuctile metals deform readily by plastic flow at low stresses, which means they require

large strains before fracture will occur, e.g. pure copper and mild steel (at room temperature).If a crack is present, it will grow when the stress is sufficiently high, leading to fracture, but

with the absorption of a large amount of energy, by ductile rupture, which is different to fastfracture. Alloys contain fine inclusions or impurities. Cracks and impurities both concentratethe stress and produce triaxial tension.

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Plastic flow occurs, producing cavities around the inclusions. These cavities link and thecrack advances by ductile tearing, which absorbs much energy per unit volume. The energyabsorbed also increases as the plastic zone size, ry, increases. Therefore materials with lowy have a large plastic zone size and absorb much energy before fracture, i.e. ductilematerials have a high fracture toughness, Kc.

Fast Fracture - CleavageBrittle materials have relatively flat featureless surfaces, e.g. ceramics. Cracks grow

rapidly, with little or no plastic flow. This is because plastic flow is much more difficult than inmetals. Also, as y is high, ry is very small, and little or no crack tip blunting occurs due to theplastic zone, i.e. the cracks remain very sharp.

Sharp cracks cause very high local:

Even with a small degree of blunting, the ideal strength of the material can be exceededat the crack tip, ie. atomic bonds break to create an atomically flat surface by cleavage.

Ductile to Brittle TransitionAt low temperatures, BCC and HCP metals are brittle (undergo fracture by cleavage), yet

they are tough above room temperature. Dislocation motion is assisted by thermal agitation.At low temperatures, dislocation motion is more difficult, so as temperature decreases, yincreases and the plastic zone size decreases. Eventually, the mechanism changes fromductile tearing to cleavage.

In steels, the ductile-brittle transition temperature is between -70C and -40C. Polymersalso have a ductile-brittle transition temperature at the glass-rubber transition, TG.

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Toughening in CompositesThe Gc of composites is greater than the Gc of polymers. In composites the Gc is

increased by reinforcement - the fibres act as crack stoppers. The mechanism is that beforethe crack reaches the fibre, the local results in debonding (where fibres lose bonding with thematrix). This causes crack blunting and arrests the motion of the crack. (However thismechanism will only be successful if the crack is perpendicular to the fibres). Thus fibres in

composites create high toughness and high stiffness.

Rubber-Toughened PolymersThe Gc of polymers may be increased by 'filler' particles. Rubber-toughened polymers use

rubber particles, e.g. ABS. The mechanism here is that the crack intersects and stretches therubber particles, which clamp the crack shut. The load required for crack motion is thereforeincreased, which increases Gc.

Avoiding Fast FractureMetals are the most important materials for highly stressed applications because of their

fracture toughness. Pure metals have a high G c above room temperature, but alloying usuallydecreases Gc, since alloying increases the lattice resistance to dislocations, i.e. y increasesand therefore ry decreases.

Practical considerations:

the extent of strengthening must be balanced with the loss of toughness

avoid brittle phases in microstructures, because they create easy crack paths.

Common examples of where easy crack paths are produced are: formation of brittle plate-like phases such as sigma phase in stainless steels or

graphite in cast irons

formation of brittle continuous grain boundary films, such as Fe 3C in annealed high

carbon steels

formation of brittle matrix structures, such as martensite in as-quenched steels,

before tempering

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MATS1101 - Materials - Fatigue Failure

Fatigue Failure

Fast Fracture and FatigueAs we have seen, a crack becomes unstable if K=Kc, and this leads to fast fracture.

Conversely, if K is less than Kc, the crack is stable and therefore, if the maximum crack size isknown, a safe load can be calculated and if the maximum load is known, the maximum crack

size allowable can be calculated.However, cracks may form and grow at lower loads if the stress is cyclic. Eventually the

crack reaches the critical size for that stress and catastrophic fast fracture will occur.

Fatigue of Uncracked Components

Fatigue tests are used to determine the likelihood of fatigue failure. Tests are carried outusing cyclic stresses, either in tension-compression or in rotating bending.

Three stress parameters are defined for the fatigue test:

Stress range

Mean stress m

Stress amplitude a

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High Cycle FatigueFor high cycle fatigue, neither max nor |min| are above the yield stress and m = 0. Under

these conditions, Basquin's law holds:

where Nf = no. of cycles to failurea = constant (0.07


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Miner's RuleWhen varies through the lifetime of the component, the cumulative damage is

summed, according to Miner's Rule. Regions, i, are defined for each different Ds, each regionuses a fraction of the life, Ni/Nfi. Failure occurs when:

where Ni = no. of cycles in region iNfi = no. of cycles to failure in region i

Fatigue of Cracked ComponentsLarge structures always contain cracks. To find the safe life of the

structure, we must know how many cycles there will be before one of thesecracks grows to the critical size for fast crack propagation. This is done byfatigue testing of pre-cracked test pieces.

A cyclic load is applied to a pre-cracked test piece. The rate of crackgrowth per cycle is measured. We define a number of parameters:

Stress intensity range, K

Mean stress intensity, Km

Stress intensity amplitude, Ka

Stress intensity range is defined as:

K increases with time at constant load, because the crack grows. Above a threshold K,the crack growth per cycle is:

where A and m are material constants

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If the initial crack length, a0, is known and the final crack length, af, at which the crackbecomes unstable is calculated using the fracture toughness K C and the maximum stressmax, then the safe number of cycles, N f, can be estimated from:

where

Fatigue Mechanisms

Pre-Cracked StructuresFor pure metals and polymers, the tensile cycle stretches the crack tip to . The plastic

zone creates a new surface. The compressive cycle then folds the new surface and the cracktip grows by ~ , i.e. da/dN ~

For alloys, which contain inclusions, the tensile cycle stretches the crack tip as in puremetals, but the inclusions form holes in the plastic zone, increasing its size. This means that increases and the crack growth rate, da/dN, is greater than that of pure metals.

Uncracked ComponentsUnder low-cycle fatigue, the plastic flow roughens the surface

quickly and the crack nucleates at the rough surface. Initially thecrack grows on the inclined slip plane. Later, the crack grows normalto the tensile axis.

Under high cycle fatigue, max is less than y, sono general plastic flow occurs. Most of the fatiguelife is spent initiating the crack. A stressconcentration (such as at a notch or change ofsection) leads to local plastic flow (because of thelocalised increased stress), which then leads tocrack initiation.

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MATS1101 - Materials - Creep

Creep

For most metals and ceramics at room temperature, strain, which is a function of stress,i.e. = f(), is independent of time. These are known as elastic/plastic solids.

As the temperature increases, they may creep at a stress which causes no permanentdeformation at room temperature. Creep is slow, continuous deformation with time. Strain, ,

now depends on temperature, T, and time, t, as well as stress, , i.e. = f(,t,T). This is acreeping solid.

The temperature at which creep occurs is dependent on the melting temperature, T m (K).Creep starts in metals when the temperature is greater than 0.3-0.4 T m. Creep starts inceramics when the temperature is greater than 0.4-0.5 Tm. Creep is rapid at temperaturesabove 0.8 Tm (K). Creep temperature can be raised by alloying, for example, superalloysoperate at 0.6-0.8 Tm.

Most polymers are not crystalline, and therefore have no melting temperature. In thesematerials, creep is related to the glass transition temperature, T G (temperature at which Vander Waals bonds melt). At temperatures higher than TG, the polymer will creep, attemperatures less than TG, the polymer is hard and shows little creep. For many polymers, T G~ room temperature, and significant creep occurs at room temperature.

Creep TestingA sample is tested for creep under uniaxial tension at constant load and temperature. The

strain is measured against time, t. Metals, polymers and ceramics all show similar creepcurves.

There are three stages in the creep behaviour of materials:1. Primary Creep. This occurs quickly, and is usually allowed for in design, as is

elastic deformation.2. Secondary Creep. This occurs at a steady state - the strain rate, (= ss) is

constant.

3. This is usually of greatest concern in design. Tertiary Creep. Creep rateaccelerates until fracture.

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Secondary CreepA). Dependence on Stress.Strain rate, , increases as n increases as given by:

where n = constant called the creep exponent (3


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Tertiary CreepInternal cavities appear during the tertiary creep stage. As the holes grow, the effective

cross-sectional area decreases, therefore the stress increases even when the load is keptconstant. As a result, the creep rate increases. Since strain rate is proportional to n, creeprate increases faster than the stress increases.

Eventually failure occurs. The time to failure, t f, depends on both stress and temperature.Data are normally given as creep-rupture diagrams:

Creep Resistant MaterialsFor a material to be resistant to creep:

It should have a high melting temperature

It should be used below 0.3Tm

Alloying can increase creep temperature to substantially above 0.3Tm

E.g. Ni-base superalloys in aircraft gas turbines operate continuously at ~0.6 Tm orintermittently at ~0.8Tm.

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MATS1101 - Materials - Creep Mechanisms

Creep Mechanisms

Dislocation Creep (Power Law Creep)

Below the elastic limit, dislocation motion is resisted by:a) intrinsic lattice resistance andb) obstacles, such as:

If dislocations can be made to climb over these obstacles, plastic flow can occur. Mostpinned dislocations have a climb force (a component of reaction force).

Dislocations only move easily by glide:

For climb to occur, the dislocation must move out of its glide plane. Atoms must beremoved from the edge of the half plane. This occurs by diffusion: if it occurs below ~0.5T m itoccurs by core diffusion (atoms diffuse along the dislocation), if it occurs above ~0.5T m it

occurs by bulk diffusion. This enables the dislocation to climb until it is free of the obstacle.

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Once the dislocation is "unlocked" by dislocation climb, it can glide, producing slip, until itmeets the next obstacle, which pins it. The climb process is then repeated, and thereforedislocation creep is progressive and continuous.

Effect of TemperatureClimb occurs by diffusion, and diffusion rate increases as temperature increases.

Therefore the rate at which dislocations become unlocked increases with temperature andcreep rate increases with increasing temperature.

Effect of StressIncreased stress results in increased driving force for dislocation climb, therefore more

dislocations climb per second and more dislocations glide per second. Therefore the creeprate increases with increasing stress.

Resisting Power Law Creep at High Stresses1. Choose a material with a high melting point (these have a low diffusion rate)2. Maximise obstruction to dislocations (with solute atoms, precipitates)3. Choose, when possible, solids with high inherent lattice resistance to dislocation

motion, e.g. ceramics and ordered alloys such as superalloys

Diffusional Creep (Linear-Viscous Creep)At low stress, in the secondary creep domain, an alternative mechanism takes place. The

observed creep exponent, n~ 1.

Creep occurs by atoms diffusing from the sides of grains to the ends to relieve the appliedstress, i.e. bulk diffusion. Grain boundary sliding is also necessary for this creep to occur.Creep rate increases as the stress increases, as the temperature increases and as the grainsize decreases.

Resisting Diffusional Creep at Low Stresses1. Choose a material with a high melting point (these have a low diffusion rate)2. Choose materials with a coarse grain size (these have less grain boundary sliding)3. Choose materials with precipitates along the grain boundaries (these impede grain

boundary sliding).

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Creep Mechanisms for PolymersFor temperatures greater than the glass transition temperature, TG, polymer chain sliding

occurs, giving Newtonian viscous flow, which is a type of creep. The creep rate increaseslinearly with stress and increases exponentially with temperature:

where C, Q & R are constantsQ is the activation energy for viscous flow (the energy required to push a lump inone molecule past that in another)

For temperatures much less than TG, polymers are elastic solids and show no creep.

Many polymers are used at temperatures around the glass transition temperature, wheremost show a mixed visco-elastic behaviour. This can be modelled by coupled spring anddashpot elements (rheology). Applying a load will lead to creep, but at an ever-decreasingrate because the spring takes up the tension. Releasing the load will lead to slow reversecreep due to the extended spring.

Resisting Polymer Creep1. Choose polymers with a high TG. TG increases with increased molecular weight,

increased cross-linking (e.g. epoxies vs. polyethylene) and increased crystallinity(high density polyethylene vs. low density polyethylene)

2. Choose polymers with fillers like silica or glass, because the creep rate decreases asthe volume fraction of filler increases (e.g. PTFE (Teflon) and polypropylene in autos)

3. Choose materials that are a polymer matrix reinforced with strong continuous fibresbecause most of the load is carried by the fibres and hence the creep resistanceincreases greatly e.g. GFRP (glass fibre reinforced plastic) and CFRP (carbon fibrereinforced plastic) composites.

materials notes

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